Question

Radioactive decay is a first order process. The most commonly used isotope in human scintigraphical procedures is 99mTc, which has a half- life of 6.0058 hours. A person was injected with 99mTc methylene phosphonate containing 13.2 × 10-6 g of 99mTc to image bone tissue and an image was obtained 75.0 minutes (1.25 h) later. What is the concentration (molarity) of 99mTc remaining at this time? ( 99mTc has a molar mass of 98.906 g /mol. Assume that the person has a blood volume of 4.80 litres).

Answer 1

Solution :-

Half life = 6.0058 h

Time passed = 1.25 h

Amount 99mTC injected = 13.2*10^-6 g

Lets first calculate the rate constant using the half life

K=0.693/ half life

   = 0.693 /6.0058 h

   = 0.11539 hr-1

Using the rate constant K we can find the amount of the 9mTC left after 1.25 hr

Ln[A]t/[A]o = - K*t

Ln[A]t/13.2*10^-6 g = - 0.11539 hr-1 * 1.25 hr

Ln[A]t/13.2*10^-6 g= -0.1442

[A]t/13.2*10^-6 g = anti ln [-0.1442]

[A]t/13.2*10^-6 g =0.8657

[A]t= 13.2*10^-6 g * 0.8657

[A]t= 1.14*10^-5 g

Now using the mass we can find the moles of 99mTC and then calculate its concentration

Moles = mass / molar mass

          = 1.14*10^-5 g / 98.906 g per mol

          = 1.15*10^-7 mol

Molarity = moles / volume in liter

                 = 1.15*10^-7 mol / 4.80 L

                 = 2.40*10^-8 M

Therefore the molarity of the 99mTc is 2.40*10^-8 M